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CBSE Class 12 Maths 2022 Question Paper 65/3/2 (Term 2) with Solutions

All 18 questions from the CBSE Class 12 Mathematics board paper, Set 65/3/2 (2022, Term 2), with answers and step-by-step solutions. Total 40 marks. Tap “Show answer & solution” under any question.

Set 65/1/1 Set 65/1/2 Set 65/1/3 Set 65/2/1 Set 65/2/2 Set 65/2/3 Set 65/3/1 Set 65/3/2 Set 65/3/3 Set 65/4/1 Set 65/4/2 Set 65/4/3 Set 65/5/1 Set 65/5/2 Set 65/5/3

Q12 marksVery Short AnswerVector Algebra

Write the projection of the vector $(b + c)$ on the vector $a$ , where $a = 2 i^- 2 j^+ k^$ , $b = i^+ 2 j^- 2 k^$ and $c = 2 i^- j^+ 4 k^$ .

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Answer: 2

$b + c = 3 i^+ j^+ 2 k^$
$a \cdot (b + c) = 6 - 2 + 2 = 6$
$∣ a ∣ = 4 + 4 + 1 = 3$
Projection $= \frac{6}{3} = 2$

Also asked in: 2022 65/3/1, 2022 65/3/3

Q22 marksVery Short AnswerDifferential Equations

Find the general solution of the differential equation : $lo g (\frac{d y}{d x}) = a x + b y$ .

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Answer:

\frac{e ^{a x}}{a} + \frac{e ^{- b y}}{b} = C

$\frac{d y}{d x} = e^{a x + b y} = e^{a x} e^{b y}$
Separate: $e^{- b y} d y = e^{a x} d x$
Integrate: $- \frac{e ^{- b y}}{b} = \frac{e ^{a x}}{a} + C_{1}$
So $\frac{e ^{a x}}{a} + \frac{e ^{- b y}}{b} = C$ ( $a, b$
eq 0 $)$

Q32 marksVery Short AnswerProbabilityNot in current syllabus

Two cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the probability distribution of the number of spade cards.

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Answer: X = 0, 1, 2 with P(X) =

\frac{9}{16}

\frac{6}{16}

\frac{3}{8}

\frac{1}{16}

Let X = number of spades; X = 0, 1, 2
P(spade) $= \frac{13}{52} = \frac{1}{4}$ , P(not spade) $= \frac{3}{4}$ , draws independent (with replacement)
$P (X = 0) = \frac{3}{4} \cdot \frac{3}{4} = \frac{9}{16}$
$P (X = 1) = 2 \cdot \frac{1}{4} \cdot \frac{3}{4} = \frac{6}{16} = \frac{3}{8}$
$P (X = 2) = \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16}$

Also asked in: 2022 65/3/1, 2022 65/3/3

Q42 marksVery Short AnswerProbability

A pair of dice is thrown and the sum of the numbers appearing on the dice is observed to be 7. Find the probability that the number 5 has appeared on atleast one die.

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Answer:

\frac{1}{3}

F: sum is 7 = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}, 6 outcomes
E: 5 on at least one die; $E \cap F$ = {(2,5), (5,2)}, 2 outcomes
$P (E ∣ F) = \frac{n ( E \cap F )}{n ( F )} = \frac{2}{6} = \frac{1}{3}$

Also asked in: 2022 65/3/1, 2022 65/3/3

Q4 (OR) (OR)2 marksVery Short AnswerProbability

The probability that A hits the target is $\frac{1}{3}$ and the probability that B hits it, is $\frac{2}{5}$ . If both try to hit the target independently, find the probability that the target is hit.

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Answer:

\frac{3}{5}

P(A misses) $= \frac{2}{3}$ , P(B misses) $= \frac{3}{5}$
P(both miss) $= \frac{2}{3} \cdot \frac{3}{5} = \frac{2}{5}$ (independent)
P(target is hit) $= 1 - \frac{2}{5} = \frac{3}{5}$

Also asked in: 2022 65/3/1, 2022 65/3/3

Q52 marksVery Short AnswerThree Dimensional GeometryNot in current syllabus

If the distance of the point (1, 1, 1) from the plane $x - y + z + λ = 0$ is $\frac{5}{3}$ , find the value(s) of $λ$ .

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Answer:

λ = 4

λ = - 6

Distance $= \frac{∣1 - 1 + 1 + λ ∣}{1 + 1 + 1} = \frac{∣1 + λ ∣}{3}$
$\frac{∣1 + λ ∣}{3} = \frac{5}{3} \Rightarrow ∣1 + λ ∣ = 5$
$1 + λ = \pm 5 \Rightarrow λ = 4$ or $λ = - 6$

Also asked in: 2022 65/3/1, 2022 65/3/3

Q62 marksVery Short AnswerIntegrals

Find : $\int \frac{d x}{x ^{2} - 6 x + 13}$

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Answer:

\frac{1}{2} tan^{- 1} (\frac{x - 3}{2}) + C

$x^{2} - 6 x + 13 = (x - 3)^{2} + 2^{2}$
$\int \frac{d x}{( x - 3 ) ^{2} + 2 ^{2}} = \frac{1}{2} tan^{- 1} (\frac{x - 3}{2}) + C$

Also asked in: 2022 65/3/1, 2022 65/3/3

Q73 marksShort AnswerVector Algebra

If $a$ , $b$ , $c$ are three vectors such that $a \cdot b = a \cdot c$ and $a \times b = a \times c$ , $a \neq = 0$ , then show that $b = c$ .

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Answer: Proved.

From $a \cdot b = a \cdot c$ : $a \cdot (b - c) = 0$ , so $b - c = 0$ or $a ⊥ (b - c)$
From $a \times b = a \times c$ : $a \times (b - c) = 0$ , so $b - c = 0$ or $a ∥ (b - c)$
A non-zero vector $b - c$ cannot be both perpendicular and parallel to the non-zero vector $a$
Hence $b - c = 0$ , i.e. $b = c$

Q7 (OR) (OR)3 marksShort AnswerVector Algebra

If $∣ a ∣ = 3$ , $∣ b ∣ = 5$ , $∣ c ∣ = 4$ and $a + b + c = 0$ , then find the value of $(a \cdot b + b \cdot c + c \cdot a)$ .

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Answer:

- 25

$(a + b + c) \cdot (a + b + c) = 0$
$∣ a ∣^{2} + ∣ b ∣^{2} + ∣ c ∣^{2} + 2 (a \cdot b + b \cdot c + c \cdot a) = 0$
$9 + 25 + 16 + 2 S = 0$
$S = - 25$

Q83 marksShort AnswerIntegrals

Evaluate : $\int_{- 1}^{2} ∣ x^{3} - x ∣ d x$

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Answer:

\frac{11}{4}

$x^{3} - x = x (x - 1) (x + 1)$ : $\geq 0$ on $[- 1, 0]$ , $\leq 0$ on $[0, 1]$ , $\geq 0$ on $[1, 2]$
Let $F (x) = \frac{x ^{4}}{4} - \frac{x ^{2}}{2}$ : $F (- 1) = - \frac{1}{4}$ , $F (0) = 0$ , $F (1) = - \frac{1}{4}$ , $F (2) = 2$
$\int_{- 1}^{0} (x^{3} - x) d x = \frac{1}{4}$
$\int_{0}^{1} (x - x^{3}) d x = \frac{1}{4}$
$\int_{1}^{2} (x^{3} - x) d x = 2 + \frac{1}{4} = \frac{9}{4}$
Total $= \frac{1}{4} + \frac{1}{4} + \frac{9}{4} = \frac{11}{4}$

Q93 marksShort AnswerThree Dimensional GeometryNot in current syllabus

Show that the lines :
$\frac{1 - x}{2} = \frac{y - 3}{4} = \frac{z}{- 1}$ and $\frac{x - 4}{3} = \frac{2 y - 2}{- 4} = z - 1$ are coplanar.

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Answer: Proved.

Line 1: $\frac{x - 1}{- 2} = \frac{y - 3}{4} = \frac{z - 0}{- 1}$ , through $(1, 3, 0)$ , d.r. $- 2, 4, - 1$
Line 2: $\frac{x - 4}{3} = \frac{y - 1}{- 2} = \frac{z - 1}{1}$ , through $(4, 1, 1)$ , d.r. $3, - 2, 1$
Condition: $4 - 1 - 2 3 1 - 3 4 - 2 1 - 0 - 1 1 = 3 - 2 3 - 2 4 - 2 1 - 1 1$
The first and third rows are equal, so the determinant is 0
Hence the lines are coplanar. (In fact $(1, 3, 0)$ lies on line 2, so they intersect there.)

Also asked in: 2022 65/3/1

Q103 marksShort AnswerDifferential Equations

Find the particular solution of the differential equation $x \frac{d y}{d x} - y = x^{2} \cdot e^{x}$ , given $y (1) = 0$ .

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Answer:

y = x e^{x} - e x

Divide by x: $\frac{d y}{d x} - \frac{1}{x} y = x e^{x}$ , linear with $P = - \frac{1}{x}$ , $Q = x e^{x}$
I.F. $= e^{- \int \frac{1}{x} d x} = e^{- l o g x} = \frac{1}{x}$
$y \cdot \frac{1}{x} = \int x e^{x} \cdot \frac{1}{x} d x = e^{x} + C$
$y = x e^{x} + C x$
$y (1) = 0$ : $0 = e + C \Rightarrow C = - e$
$y = x e^{x} - e x$

Also asked in: 2022 65/3/1, 2022 65/3/3

Q10 (OR) (OR)3 marksShort AnswerDifferential Equations

Find the general solution of the differential equation $x \frac{d y}{d x} = y (lo g y - lo g x + 1)$ .

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Answer:

lo g (\frac{y}{x}) = C x

$\frac{d y}{d x} = \frac{y}{x} (lo g \frac{y}{x} + 1)$ , a homogeneous equation
Put $y = v x$ : $v + x \frac{d v}{d x} = v (lo g v + 1)$
$x \frac{d v}{d x} = v lo g v$
$\int \frac{d v}{v l o g v} = \int \frac{d x}{x}$
$lo g ∣ lo g v ∣ = lo g ∣ x ∣ + lo g ∣ C ∣$
$lo g v = C x$ , i.e. $lo g (\frac{y}{x}) = C x$

Also asked in: 2022 65/3/1, 2022 65/3/3

Q114 marksLong AnswerIntegrals

Find : $\int \frac{x ^{2}}{( x ^{2} + 1 ) ( 3 x ^{2} + 4 )} d x$

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Answer:

- tan^{- 1} x + \frac{2}{3} tan^{- 1} (\frac{3 x}{2}) + C

Put $x^{2} = t$ : $\frac{t}{( t + 1 ) ( 3 t + 4 )} = \frac{A}{t + 1} + \frac{B}{3 t + 4}$
$t = A (3 t + 4) + B (t + 1)$ ; $t = - 1$ : $A = - 1$ ; $t = - \frac{4}{3}$ : $B = 4$
So integrand $= - \frac{1}{x ^{2} + 1} + \frac{4}{3 x ^{2} + 4}$
$\int \frac{4}{3 x ^{2} + 4} d x = \frac{4}{3} \int \frac{d x}{x ^{2} + ( \frac{2}{3} ) ^{2}} = \frac{4}{3} \cdot \frac{3}{2} tan^{- 1} \frac{3 x}{2} = \frac{2}{3} tan^{- 1} \frac{3 x}{2}$
Answer: $- tan^{- 1} x + \frac{2}{3} tan^{- 1} (\frac{3 x}{2}) + C$

Also asked in: 2022 65/3/1, 2022 65/3/3

Q11 (OR) (OR)4 marksLong AnswerIntegrals

Evaluate : $\int_{- 2}^{1} 5 - 4 x - x^{2} d x$

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Answer:

\frac{9 π}{4}

$5 - 4 x - x^{2} = 9 - (x + 2)^{2}$
$\int 9 - (x + 2)^{2} d x = \frac{x + 2}{2} 5 - 4 x - x^{2} + \frac{9}{2} sin^{- 1} \frac{x + 2}{3}$
At $x = 1$ : $\frac{3}{2} \cdot 0 + \frac{9}{2} \cdot \frac{π}{2} = \frac{9 π}{4}$
At $x = - 2$ : $0 + 0 = 0$
Value $= \frac{9 π}{4}$

Also asked in: 2022 65/3/1, 2022 65/3/3

Q124 marksLong AnswerApplication of Integrals

Using integration, find the area of the region bounded by the curves $x^{2} + y^{2} = 4$ , $x = 3 y$ and $x$ -axis lying in the first quadrant.

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Answer:

\frac{π}{3}

sq units

Intersection of $y = \frac{x}{3}$ with $x^{2} + y^{2} = 4$ : $x^{2} + \frac{x ^{2}}{3} = 4 \Rightarrow x = 3$ , $y = 1$
Area $= \int_{0}^{3} \frac{x}{3} d x + \int_{3}^{2} 4 - x^{2} d x$
$\int_{0}^{3} \frac{x}{3} d x = \frac{1}{3} \cdot \frac{3}{2} = \frac{3}{2}$
$\int_{3}^{2} 4 - x^{2} d x = [\frac{x}{2} 4 - x^{2} + 2 sin^{- 1} \frac{x}{2}]_{3}^{2} = π - (\frac{3}{2} + \frac{2 π}{3}) = \frac{π}{3} - \frac{3}{2}$
Total area $= \frac{π}{3}$ sq units

Q134 marksLong AnswerThree Dimensional GeometryNot in current syllabus

Find the distance of the point $(1, - 2, 9)$ from the point of intersection of the line $r = 4 i^+ 2 j^+ 7 k^+ λ (3 i^+ 4 j^+ 2 k^)$ and the plane $r \cdot (i^- j^+ k^) = 10$ .

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Answer: 10 units

General point of the line: $(4 + 3 λ, 2 + 4 λ, 7 + 2 λ)$
On the plane $x - y + z = 10$ : $(4 + 3 λ) - (2 + 4 λ) + (7 + 2 λ) = 10$
$9 + λ = 10 \Rightarrow λ = 1$ , point $(7, 6, 9)$
Distance from $(1, - 2, 9)$ $= 6^{2} + 8^{2} + 0^{2} = 10$ units

Also asked in: 2022 65/3/1, 2022 65/3/3

Q144 marksCase StudyProbability

A shopkeeper sells three types of flower seeds A1, A2, A3. They are sold in the form of a mixture, where the proportions of these seeds are 4 : 4 : 2, respectively. The germination rates of the three types of seeds are 45%, 60% and 35% respectively.
Based on the above information :
(a) Calculate the probability that a randomly chosen seed will germinate; (2)
(b) Calculate the probability that the seed is of type A2, given that a randomly chosen seed germinates. (2)

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Answer: (a) 0.49 (b)

\frac{24}{49}

$P (A_{1}) = 0.4$ , $P (A_{2}) = 0.4$ , $P (A_{3}) = 0.2$ ; G = seed germinates
$P (G ∣ A_{1}) = 0.45$ , $P (G ∣ A_{2}) = 0.60$ , $P (G ∣ A_{3}) = 0.35$
(a) $P (G) = 0.4 \times 0.45 + 0.4 \times 0.60 + 0.2 \times 0.35 = 0.18 + 0.24 + 0.07 = 0.49$
(b) By Bayes' theorem, $P (A_{2} ∣ G) = \frac{0.24}{0.49} = \frac{24}{49}$

Also asked in: 2022 65/3/1, 2022 65/3/3

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